Question

will give medal 3 questions help please,
1. Suppose y varies inversely with x. Write an equation for the inverse variation. Y = 5 when x = -5

2. Graph each inverse variation. Y = -18 / x

3.Identify the excluded value for each function f(x)=5/x

Answer 1

1. y = k/x
4 = k/-6
k = -24
y = -24/x
x = -24/y

Answer 2

are you sure this is current ?

Answer 3

most likely yes

Answer 4

can you do the others please

Answer 5

3.

Excluded values occur to prohibit division by 0 (zero).
f(x) = 5/x then this function is meaningless when x = 0 and it is said to be discontinuous at x = 0.
Take your denominator and set it not equal to zero, since dividing by zero would give you an answer that is undefined. In the case of this question the denominator is x, therefore x=/=0, since a x value equal to 0 would cause the denominator to be zero.

Answer 6

okay thank you so much what about number 2 ?

Answer 7

sorry can't help u with that

Answer 8

For number 2 just plug that into your graphing calc if you dont have one use wolframalpha :)

Answer 9

@tester97 your right i totally forgot about wolfram lolll thanks

Answer 10

I did it for you already so here ya go enjoy :) http://www.wolframalpha.com/input/?i=Y+%3D+-18+%2F+x

Answer 11

@ryanvarghese12790 i have two more
Identify the asymptotes of the graph of each function. Then graph the function. y=1/x+2

Identify the asymptotes of the graph of each function. Then graph the function. Y = 5 / x – 4 + 1

Answer 12

omgggg @tester97 thank you thats correct i wish i could give you a medal

Answer 13

welcome |:) just use wolframalpha to graph these last two ^_^

Answer 14

ohhhh your right i can well then i have a word problem if you guys can help

Answer 15

sorry can't help u with the other two q's, this is not one of my strongest topics, I am terribly sorry for the inconvenience this has caused @bettylee

Answer 16

its okay you helped me with the others which i very much appreciated loll you dont have to apologized you actually helped a lot. but here is a word problem

1.for a 225-watt bulb, the intensity i of light of humens at a distance of x feet is i=225/x^2
A.what is the intensity of light 5ft from the bulb
B. suppose your distance from the bulb doubles, how does the intensity of the light change ? explain

Answer 17

>.< sorry im not good with word probz

Answer 18

okay loll well i wasd talking to ryan lolll hopefully he can help, ummm can you graph this on wolfram for me Y = 5 / x – 4 + 1

Answer 19

http://www.wolframalpha.com/input/?i=Y+%3D+5+%2F+x+%E2%80%93+4+%2B+1

Answer 20

thankssss @tester97 you rock! ( :